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Mathematics Question on integral

Integrate the function: 1−x1+x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}

Answer

I=1−x1+xI=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}
Let  x=cos2θ⇒dx=−2sinθcosθ  dθLet\space x=cos^{2}θ⇒dx=-2sinθcosθ\space dθ
I=∫1−cosθ1+cosθ(−2sinθcosθ)dθI=\int\sqrt{\frac{1-cosθ}{1+cosθ}}(-2sinθcosθ)dθ
=−∫2sin2θ22cos2θ2sin2θdθ=-\int\sqrt{\frac{2sin^{2}\frac{θ}{2}}{2cos^{2}\frac{θ}{2}}}sin2θdθ
=−∫tanθ2.sinθcosθdθ=-\int tan\frac{θ}{2}.sinθcosθdθ
=−2∫sinθ2cosθ2(2sinθ2cosθ2)cosθdθ=-2\int \frac{sin\frac{θ}{2}}{cos\frac{θ}{2}}(2sin\frac{θ}{2}cos\frac{θ}{2})cosθdθ
=−4∫sin2θ2cosθdθ=-4\int sin^{2}\frac{θ}{2}cosθdθ
=−4∫sin2θ2.(2cos2θ2−1)dθ=-4\int sin^{2}\frac{θ}{2}.(2cos^{2}\frac{θ}{2}-1)dθ
=−4∫(2sin2θ2cos2θ2−sin2θ2)dθ=-4\int (2sin^{2}\frac{θ}{2}cos^{2}\frac{θ}{2}-sin^{2}\frac{θ}{2})dθ
=−8∫sin2θ2.cos2θ2dθ+4∫sin2θ2dθ=-8\int sin^{2}\frac{θ}{2}.cos^{2}\frac{θ}{2}dθ+4\int sin^{2}\frac{θ}{2}dθ
=−2∫sin2θdθ+4∫sin2θ2dθ=-2\int sin^{2}θdθ+4\int sin^{2}\frac{θ}{2}dθ
=−2∫(1−cos2θ2)dθ+4∫1−cosθ2dθ=-2\int(\frac{1-cos2θ}{2})dθ+4\int\frac{1-cosθ}{2}dθ
=−2[θ2−sin2θ4]+4[θ2−sinθ2]+C=−θ+sin2θ2+2θ−2sinθ+C=-2[\frac{θ}{2}-\frac{sin2θ}{4}]+4[\frac{θ}{2}-\frac{sinθ}{2}]+C =-θ+\frac{sin2θ}{2}+2θ-2sinθ+C
=θ+sin2θ2−2sinθ+C=θ+\frac{sin2θ}{2}-2sinθ+C
=θ+2sinθcosθ2−2sinθ+C=θ+\frac{2sinθcosθ}{2}-2sinθ+C
=θ+1−cos2θ.cosθ−21−cos2θ+C=θ+\sqrt{1-cos^{2}θ}.cosθ-2\sqrt{1-cos^{2}θ}+C
=cos−1x+1−x.x−21−x+C=cos^{-1}\sqrt{x}+\sqrt{1-x}.\sqrt{x}-2\sqrt{1-x}+C
=−21−x+cos−1x+x(1−x)+C=-2\sqrt{1-x}+cos^{-1}\sqrt{x}+\sqrt{x(1-x)}+C
=−21−x+cos−1x+x−x2+C=-2\sqrt{1-x}+cos^{-1}\sqrt{x}+\sqrt{x-x^{2}}+C